[SFW] Riddles / Puzzles / Brain Teasers - Page 17

No; sorry, you're missing the point.

You say that the counter adds 1 if the left lever is up, and flips it down.
You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers.
Everyone else flips the left lever up the first chance they get, but not subsequently.

Suppose the left lever starts down.
A non-counter enters the room first (unknowingly) and flicks the left lever up.
Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.

Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.

They tell their salary to a sixth person who they don´t work with, meaning he´s not a coworker.

Let's call these people a, b and c with salaries A, B and C.

Then a needs to find out (B + C) without ever being able to individually deduce B or C, b needs to find out (A + C), etc.

Then b pretty much has to give some altered (but undecipherable) form of his salary to c, so c can pass the (altered) sum on to a. The complete sum has to come back to b to be deciphered, since otherwise one of the others would be able to use the information to deduce his salary.

So, let's have a, b and c generate some arbitrary numbers and hand them out.
- a thinks of a number D and tells b only
- b thinks of a number E and tells c only
- c thinks of a number F and tells a only

Then each person multiplies their salary by the number and gives it to the third person:
- a tells AF to b (who doesn't know F)
- b tells BD to c (who doesn't know D)
- c tells CE to a (who doesn't know E)

Now we can give salted sums of 2 salaries for the 3rd person to add his. Although the 3rd person knows the multiplier, he's given a sum of 2 salaries so can cannot work out either.
- a tells (A + CE) to b
- b tells (B + AF) to c
- c tells (C + BD) to a

We can now get people to put together all 3 salaries, and tell this sum to the person with the modified salary in the sum to decipher:
- a tells (A + BD + C) to b
- b tells (A + B + CE) to c
- c tells (AF + B + C) to a

Now each of the people can subtract (their own salary * the multiplier they were given at the start), add their own salary, and divide by 3 to get the average.

At no point is anyone able to calculate either of the other two people's salaries.

Does this same thing work for 5 people?

Let's call these people a, b, c, d and e with salaries A, B, C, D and E.

So, let's have a, b and c generate some arbitrary numbers and hand them out.
- a thinks of a number F and tells b only
- b thinks of a number G and tells c only
- c thinks of a number H and tells d only
- d thinks of a number I and tells e only
- e thinks of a number J and tells a only

Then each person multiplies their salary by the number and gives it to the next person:
- a tells AJ to b (who doesn't know J)
- b tells BF to c (who doesn't know F)
- c tells CG to d (who doesn't know G)
- d tells DH to e (who doesn't know H)
- e tells EI to a (who doesn't know I)

Now we can give salted sums of 2 salaries for a 3rd person to add his.
- a tells (A + EI) to b
- b tells (B + AJ) to c
- c tells (C + BF) to d
- d tells (D + CG) to e
- e tells (E + DH) to a

And again each person adds their salary and passes it around again, for a 4th person to add his.
- a tells (A + DH + E) to b
- b tells (A + B + EI) to c
- c tells (AJ + B + C) to d
- d tells (BF + C + D) to e
- e tells (CG + D + E) to a

And again.
- a tells (A + CG + D + E) to b
- b tells (A + B + DH + E) to c
- c tells (A + B + C + EI) to d
- d tells (AJ + B + C + D) to e
- e tells (BF + C + D + E) to a

One final time, to pass the final sum around. The final sum arrives at the person whose own salary is altered, so they can decipher it.
- a tells (A + BF + C + D + E) to b
- b tells (A + B + CG + D + E) to c
- c tells (A + B + C + DH + E) to d
- d tells (A + B + C + D + EI) to e
- e tells (AJ + B + C + D + E) to a

Now each of the people can subtract (their own salary * the multiplier they were given at the start), add their own salary, and divide by 5 to get the average.

At no point is anyone able to calculate any other the other people's salaries.

Thinking about it, was it completely unnecessary to have someone else generate a multiplier in the first step?

All we actually need is for each person to generate a random number, get everyone else to add their salary to it in turn, and then when it gets back to each person only they can subtract their random number and add their salary to get the total :p

The simplest way is to find a trusted 3rd party who promises not to tell anyone what anyone else's salary is, have everyone tell that person their salary, and they work out the average and let everyone know what it is. But if you can't find such a person, there's another way...

Arrange the cow-orkers in a line, and have them all pick a random number and remember it (without telling anyone else). This number is allowed to be negative, and can be as big as they like.

The first person adds his salary and random number together, and tells that sum to the next person (without telling anyone else). They add on their salary and their random number, and tell the new total to the next person, who does the same, through all of them.

The last person, after adding on her salary and random number, tells her result to the first person.

The first person then subtracts his random number from the number he was told, and passes the answer on to the second person. They subtract their random number, pass the result on, down the line again. The last person, after subtracting her random number, announces the number to everybody. That is the total salary of them all, and they just have to divide by the number of people to get the average.

For the first person, the only information they get passed is the sum of everyone's salary, obscured by all their random numbers. He can't figure anything out from that.

For everyone else except the last person, the two numbers they are told are obscured either by the random numbers of the people in front of them (the first time through) or after them (the second time through). So they can't tell anything either.

The last person has nobody after them, so the number they get the second time through tells them something! Well no, not really, since all it tells them is the total salary of everyone else, and everyone else can work out the same thing once they are told the total salary, just by subtracting their own salary. So the system does not reveal any additional information about anyone else's salary to anyone. QED

(however, the system is not perfect. If the people standing on either side of someone get together, they can work out that person's salary without revealing to each other anything more about their own salaries.)

9 races. I'll break them into lettered groups.

A: 5 races of 5 horses each. Group stages.
B: 1 race of 5 winners of group stages A. Winner is your fastest horse.
C: 1 race of runner ups of A.
D: 1 race of 3rd place finishers of A.
E: 1 race to determine 2nd/3rd fastest horses:
-2nd and 3rd place of race B
-1st and 2nd place of race C
-1st place of race D

Rationale:
-Race B winner is fastest horse, so you only have to worry about the other 2 slots.
-Race D: to qualify, the horse finished 3rd in its original group, meaning 2 horses were already faster. Therefore only the winner in Race D has a chance to be among the top 3 overall (or it'd be at best 4th, behind the 1st/2nd place in its race A, and 1st in race D).
-Similar rationale for Race C: only the 1st/2nd place finishers have a chance to be top 3 overall, or else they'd be at best 4th (behind 1st place in their race A, and 1st/2nd in race C).
-You have exactly 5 potential horses remaining for race E to determine your 2nd/3rd fastest horses overall.

EDIT: This assumes you couldn't identify horses across races. If you could, it would be easier, and take 7 races.

Same A and B as above. Then one race of:
-2nd and 3rd place of race B
-2nd and 3rd place of the race A that had the horse who won race B
-2nd place of the race A that had the horse who was 2nd in race B

Reason here being if you finished 2nd/3rd in a race where the winner wasn't the top 3, there's no chance that you're one of the top 3.

You don't spin the barrel. If you spin it, you have a 1/3 chance of landing on a chamber with a bullet, and 2/3 chance of landing on a chamber without one. If you don't spin it, you have a 1/4 chance of the next chamber having a bullet, and a 3/4 chance of the next chamber being empty. 3/4 > 2/3.

Statistically, you should not spin it. The chance is 1/4 that the first shot was in the position near the 2 consecutive bullets. There is a 3/4 or 75% chance that it was in locations 1-3 of the 4 consecutive empty slots. Some people are going to argue for the slots that are filled with bullets should be taken into the equation, but knowing that the first shot was a miss means that you know you are dealing with bullet slots 1-5 as the very worst spot the bullet could be at this moment is in slot number 5. there is only a 25% chance it is there. Because you know as soon as the first shot doesnt go off that it its NOT in slot 1. Therefore it can only be slots 2-5.

If you spin the gun you chance of survival drops from 75% to 66%. Its a 6 round chamber, and 2/6 or 1/3 are bullets. You are giving an opportunity for it to land on any of the 6. Whereas before there is only 1 bullet to 3 empty slots possible.

Edited, was wrong T_T person above was right

Your rationale only applies if the chamber the gun fires from is random, but it isn't. It always fires the next bullet in the chamber.